The topology on the ring of polynomials and the continuity of the evaluation map

Polynomials are an algebraic gadget, and one is rarely led to think about the topology a ring of polynomials should carry. That happened to me, though, more or less by accident, when María Inés de Frutos Fernández and I worked on implementing in Lean the evaluation of power series. So let's start with them. To simplify the discussion, I only consider the case of one inderminate. When there are finitely many of them, the situation is the same; in the case of infinitely many indeterminates, there might be some additional subtleties, but I have not thought about it.

$\gdef\lbra{[\![}\gdef\rbra{]\!]} \gdef\lpar{(\!(}\gdef\rpar{)\!)} \gdef\bN{\mathbf N} \gdef\coeff{\operatorname{coeff}} \gdef\eval{\operatorname{eval}} \gdef\colim{\operatorname{colim}}$

Power series

A power series over a ring $R$ is just an expression $\sum a_nT^n$, where $(a_0,a_1, \dots)$ is a family of elements of $R$ indexed by the integers. After all, this is just what is meant by “formal series”: coefficients and nothing else.

Defining a topology on the ring $R\lbra T\rbra$ should allow to say what it means for a sequence $(f_m)$ of power series to converge to a power series $f$, and the most natural thing to require is that for every $n$, the coefficient $a_{m,n}$ of $T^n$ in $f_m$ converges to the corresponding coeffient $a_m$ of $T^n$ in $f$. In other words, we endow $R\lbra T\rbra$ with the product topology when it is identified with the product set $R^{\bN}$. The explicit definition may look complicated, but the important point for us is the following characterization of this topology: Let $X$ be a topological space and let $f\colon X \to R\lbra T\rbra$ be a map; for $f$ to be continuous, it is necessary and sufficient that all maps $f_n\colon X \to R$ are continuous, where, for any $x\in X$, $f_n(x)$ is the $n$th coefficient of $f(x)$. In particular, the coeffient maps $R\lbra T\rbra\to R$ are continuous.

What can we do with that topology, then? The first thing, maybe, is to observe its adequacy wrt the ring structure on $R\lbra T\rbra$.

Proposition. — If addition and multiplication on $R$ are continuous, then addition and multiplication on $R\lbra T\rbra$ are continuous.

Let's start with addition. We need to prove that $s\colon R\lbra T\rbra \times R\lbra T\rbra\to R\lbra T\rbra$ is continuous. By the characterization, it is enough to prove that all coordinate functions $s_n\colon R\lbra T\rbra \times R\lbra T\rbra\to R$, $(f,g)\mapsto \coeff_n(f+g)$, are continuous. But these functions factor through the $n$th coefficient maps: $\coeff_n(f+g) = \coeff_n(f)+\coeff_n(g)$, which is continuous, since addition, coefficients and projections are continuous. This is similar, but slightly more complicated for multiplication: if the multiplication map is denoted by $m$, we have to prove that the maps $m_n$ defined by $m_n(f,g)=\coeff_n(f\cdot g)$ are continuous. However, they can be written as $$m_n(f,g)=\coeff_n(f\cdot g) = \sum_{p=0}^n \coeff_p(f)\coeff_{n-p}(g).$$ Since the projections and the coefficient maps are continuous, it is sufficient to prove that the maps from $R^{n+1} \times R^{n+1}$ to $R$ given by $$((a_0,\dots,a_n),(b_0,\dots,b_n))\mapsto \sum_{p=0}^n a_p b_{n-p}$$ are continuous, and this follows from continuity and commutativity of addition on $R$, because it is a polynomial expression.

Polynomials

At this point, let's go back to our initial question of endowing polynomials with a natural topology.

An obvious candidate is the induced topology. This looks correct; in any case, it is such that addition and multiplication on $R[T]$ are continuous. However, it lacks an interesting property with respect to evaluation.

Recall that for every $a\in R$, there is an evaluation map $\eval_a\colon R[T]\to R$, defined by $f\mapsto f(a)$, and even, if one wishes, the two-variable evaluation map $R[T]\times R\to R$.
The first claim is that this map is not continuous.

An example will serve of proof. I take $R$ to be the real numbers, $f_n=T^n$ and $a=1$. Then $f_n$ converges to zero, because for each integer $m$, the real numbers $\coeff_m(f_n)$ are zero for $n>m$. On the other hand, $f_n(a)=f_n(1)=1$ for all $n$, and this does not converge to zero!

So we have to change the topology on polynomials if we want that this map be continuous, and we now give the correct definition. The ring of polynomials is the increasing union of subsets $R[T]_n$, indexed by integers $n$, consisting of all polynomials of degree less than $n$. Each of these subsets is given the product topology, as above, but we endow their union with the “inductive limit” topology. Explicitly, if $Y$ is a topological space and $u\colon R[T]\to Y$ is a map, then $u$ is continuous if and only if, for each integer $n$, its restriction to $R[T]_n$ is continuous.

The inclusion map $R[T]\to R\lbra T\rbra$ is continuous, hence the topology on polynomials is finer than the topology induced by the topology on power series. As the following property indicates, it is usually strictly finer.

We can also observe that addition and multiplication on $R[T]$ are still continuous. The same proof as above works, once we observe that the coefficient maps are continuous. (On the other hand, one may be tempted to compare the product topology of the inductive topologies, with the inductive topology of the product topologies, a thing which is not obvious in the direction that we need.)

Proposition. — Assume that addition and multiplication on $R$ are continuous. Then the evaluation maps $\eval_a \colon R[T]\to R$ are continuous.

We have We have to prove that for every integer $n$, the evaluation map $\eval_a$ induced a continuous map from $R[T]_n$ to $R$. Now, this map factors as a projection map $R[T]\to R^{n+1}$ composed with a polynomial map $(c_0,\dots,c_n)\mapsto c_0+c_1a+\dots+c_n a^n$. It is therefore continuous.

Laurent series

We can upgrade the preceding discussion and define a natural topology on the ring $R\lpar T\rpar$ of Laurent series, which are the power series with possibly negative exponents. For this, for all integers $d$, we set $R\lpar T\rpar_d$ to be the set of power series of the form $f=\sum_{n=-d}^\infty c_n T^n$, we endow that set with the product topology, and take the corresponding inductive limit topology. We leave to the reader to check that this is a ring topology, but that the naïve product topology on $R\lpar T\rpar$ wouldn't be in general.

Back to the continuity of evaluation

The continuity of the evaluation maps $f\mapsto f(a)$ were an important guide to the topology of the ring of polynomials. This suggests a more general question, for which I don't have a full answer, whether the two-variable evaluation map, $(f,a)\mapsto f(a)$, is continuous. On each subspace $R[T]_d\times R$, the evaluation map is given by a polynomial map ($(c_0,\dots,c_d,a)\mapsto c_0 +c_1a+\dots+c_d a^d$), hence is continuous, but that does not imply the desired continuity, because that only tells us about $R[T]\times R$ with the topology $\colim_d (R[T]_d\times R)$, while we are interested in the topology $(\colim_d R[T]_d)\times R$. To compare these topologies, note that the natural bijection $\colim_d (R[T]_d\times R) \to (\colim_d R[T]_d)\times R$ is continuous (because it is continuous at each level $d$), but the continuity of its inverse is not so clear.

I find it amusing, then, to observe that sequential continuity holds in the important case where $R$ is a field. This relies on the following proposition.

Proposition. — Assume that $R$ is a field. Then, for every converging sequence $(f_n)$ in $R[T]$, the degrees $\deg(f_n)$ are bounded.

Otherwise, we can assume that $(f_n)$ converges to $0$ and that $\deg(f_{n+1})>\deg(f_n)$ for all $n$. We construct a continuous linear form $\phi$ on $R[T]$ such that $\phi(f_n)$ does not converge to $0$. This linear form is given by a formal power series $\phi(f)=\sum a_d c_d$ for $f=\sum c_dT^d$, and we choose the coefficients $(a_n)$ by induction so that $\phi(f_n)=1$ for all $n$. Indeed, if the coefficients are chosen up to $\deg(f_n)$, then we fix $a_d=0$ for $\deg(f_n)<d<\deg(f_{n+1})$ and choose $a_{\deg(f_{n+1})}$ so that $\phi(f_{n+1})=1$. This linear form is continuous because its restriction to any $R[T]_d$ is given by a polynomial, hence is continuous.

Corollary. — If $R$ is a topological ring which is a field, then the evaluation map $R[T]\times R\to R$ is sequentially continuous.

Consider sequences $(f_n)$ in $R[T]$ and $(a_n)$ in $R$ that converge to $f$ and $a$ respectively. By the proposition, there is an integer $d$ such that $\deg(f_n)\leq d$ for all $n$, and $\deg(f)\leq d$. Since evaluation is continuous on $R[T]_d\times R$, one has $f_n(a_n)\to f(a)$, as claimed.

Remark. — The previous proposition does not hold on rings. In fact, if $R=\mathbf Z_p$ is the ring of $p$-adic integers, then $\phi(p^nT^n)=p^n \phi(T^n)$ converges to $0$ for every continuous linear form $\phi$ on $R[T]$. More is true since in that case, evaluation is continuous! The point is that in $\mathbf Z_p$, the ideals $(p^n)$ form a basis of neighborhoods of the origin.

Proposition. — If the topology of $R$ is linear, namely the origin of $R$ has a basis of neighborhoods consisting of ideals, then the evaluation map $R[T]\times R\to R$ is continuous.

By translation, one reduces to showing continuity at $(0,0)$. Let $V$ be a neighborhood of $0$ in $R$ and let $I$ be an ideal of $R$ such that $I\subset V$. Since it is an subgroup of the additive group of $R$, the ideal $I$ is open. Then the set $I\cdot R[T]$ is open because for every $d$, its trace on $R[T]_d$, is equal to $I\cdot R[T]_d$, hence is open. Then, for $f\in I\cdot R[T]$ and $a\in R$, one has $f(a)\in I$, hence $f(a)\in V$.

Here is one case where I can prove that evaluation is continuous.

Proposition. — If the topology of $R$ is given by a family of absolute values, then the evaluation map $(f,a)\mapsto f(a)$ is continuous.

I just treat the case where the topology of $R$ is given by one absolute value. By translation and linearity, it suffices to prove continuity at $(0,0)$. Consider the norm $\|\cdot\|_1$ on $R[T]$ defined by $\|f\|_1=\sum |c_n|$ if $f=\sum c_nT^n$. By the triangular inequality, one has $|f(a)|\leq \|f\|_1$ for any $a\in R$ such that $|a|\leq 1$. For every $r>0$, the set $V_r$ of polynomials $f\in R[T]$ such that $\|f\|_1<r$ is an open neighborhood of the origin since, for every integer $d$, its intersection with $R[T]_d$ is an open neighborhood of the origin in $R[T]_d$. Let also $W$ be the set of $a\in R$ such that $|a|\leq 1$. Then $V_r\times W$ is a neighborhood of $(0,0)$ in $R[T]\times R$ such that $|f(a)|<r$ for every $(f,a)\in V_r\times W$. This implies the desired continuity.